GATE CS 1987 Q2d — Sets, Relations and Functions

The given statement is FALSE (represented as 0).

For a relation to be an equivalence relation, it must be reflexive, symmetric, and transitive. While the union of two equivalence relations is always reflexive and symmetric, it is not necessarily transitive.

Counterexample:

Let $A = \{1, 2, 3\}$.

Define two equivalence relations on $A$:

• $R_1 = \{(1,1), (2,2), (3,3), (1,2), (2,1)\}$
• $R_2 = \{(1,1), (2,2), (3,3), (2,3), (3,2)\}$

Their union is:

In $R_1 \cup R_2$, we have:

• $(1,2) \in R_1 \cup R_2$
• $(2,3) \in R_1 \cup R_2$

But $(1,3) \notin R_1 \cup R_2$. Thus, the transitivity property fails, and $R_1 \cup R_2$ is not an equivalence relation.